Let $R$ be the region enclosed by the $y$ -axis, the line $y=1$ and the curve $y=x^3$. $y$ $x$ ${y=x^3}$ $ R$ $ 0$ $ 1$ $ 1$ A solid is generated by rotating $R$ about the line $y=1$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=x^3}$ Each slice is a cylinder. Let the width of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=x^3}$ $ 0$ $ 1$ $ 1$ $r$ The radius is equal to the distance between the curve $y=x^3$ and the line $y=1$. In other words, for any $x$ -value, this is the equation for $r(x)$ : $\begin{aligned} r(x)}&=1-x^3} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(1-x^3}\right)^2 \\\\ &=\pi(1-2x^3+x^6) \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=1$. So the interval of integration is $[0,1]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^1 \left[\pi (1-2x^3+x^6)\right]dx \\\\ &=\pi \int_0^1 (1-2x^3+x^6)\, dx \end{aligned}$ Let's evaluate the integral. $\pi \int_0^1 (1-2x^3+x^6)\, dx=\dfrac{9\pi}{14}$ In conclusion, the volume of the solid is $\dfrac{9\pi}{14}$.